Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder istan² α.

The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as:

Here, a cylinder of radius R and height H is inscribed in the cone.

Then, ∠GAO = α, OG = r, OA = h, OE = R, and CE = H.

We have,

r = h tan α

Now, since ΔAOG is similar to ΔCEG, we have:

Now, the volume (V) of the cylinder is given by,

And, for, we have:

∴By second derivative test, the volume of the cylinder is the greatest when

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as:

Hence, the given result is proved.